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A point moves along a curved path y=0.4x2, At x=2m its speed is 6 m/s increasing at 3 m/s2. At this instatnt find- (i) Velocity components along x and y direction. (ii) Its acceleration.
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$y= 0.4\ x^2 \cdots (Given)\ Hence,\ \dfrac {dy} {dx}= 0.8x\ and \dfrac {d^2y} {dx^2}=0.8\ At\ x=2, \dfrac {dy} {dx}= 0.8 \times 2= 1.6\ and \dfrac {d^2y} {dx^2}=0.8\$

Radius of curvature (ρ)=[1+ (dy/dx)2]3/2d2y/dx2[1+(1.6)2]3/20.8=8.3962 m.

x=2, velocity(v)=6m/s and tangential acceleration (at)=3m/s2(Given)With the horizontal, let velocity vector make an angle θHence, tanθ=m=slope of tangent=dydx=1.6

θ=58 Velocity along xdirection (vx)=vcosθ=6.cos58=3.1795 m/sVelocity along ydirection (vy)=vsinθ=6.sin58=5.0883 m/s

Normal acceleration (an)=v2ρ=628.3962=4.2877 m/s2a2=a12+a22=32+4.28772=27.3844Hence, (a)=5.2330 m/s2.

Let acceleration vector make an angle α with at.Hence, α=tan1anat=tan14.28773=55.0204 With horizontal, angle made by acceleration =θ+α=55.0204+58=113.0204

Hence,Velocity along xdirection (vx)=3.1795 m/s; Velocity along ydirection; (vy)=5.0883 m/s and acceleration=5.2330 m/s2,=113.0204 

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